Problem: N different natural numbers are given. If you select four different ones and set them as $ A $, $ B $, $ C $, $ D $, the following formula

$ \ Frac {A + B} {C --D} $



I want to find the maximum value of.




Given N different natural numbers, choose 4 different from them and create a program to find the maximum value of the above formula.



Input

The input is given in the following format.


N
a1 a2 ... aN


The number N (4 ≤ N ≤ 1000) of natural numbers is given in the first line. The value ai (1 ≤ ai ≤ 108) of each natural number is given in the second line. However, the same natural number does not appear more than once (ai ≠ aj for i ≠ j).

Output

Outputs the maximum value of the above formula as a real number for a given N natural numbers. However, the error must not exceed plus or minus 10-5.

Examples

Input

10
1 2 3 4 5 6 7 8 9 10


Output

19.00000


Input

5
22 100 42 3 86


Output

9.78947


Input

6
15 21 36 10 34 5


Output

18.00000


Input

4
100000 99999 8 1


Output

28571.285714

Solution (in CPP):

Solution:
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
using namespace std;

int main(void) {
  int i, j, k, l, n;
  double ans = 0;
  scanf("%d", &n);
  int a[n];
  pair<int, int> d[n];
  for(i = 0; i < n; ++i) scanf("%d", &a[i]);
  sort(a, a + n, greater<int>());
  for(i = 0; i < n - 1; ++i) d[i].first = a[i] - a[i + 1], d[i].second = i;
  sort(d, d + n - 1);
  if(d[0].second && d[0].second != 1) ans = (double)(a[0] + a[1]) / d[0].first;
  else {
    for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) for(k = 0; k < 4; ++k) for(l = 0; l < 4; ++l) if(i != j && j != k && k != l && l != i && i != k && j != l) if(ans < (double)(a[i] + a[j]) / (a[k] - a[l])) ans = (double)(a[i] + a[j]) / (a[k] - a[l]);
    if(d[1].second && d[1].second != 1) {
      if(ans < (double)(a[0] + a[1]) / d[1].first) ans = (double)(a[0] + a[1]) / d[1].first;
    } else {
      if(ans < (double)(a[0] + a[1]) / d[2].first) ans = (double)(a[0] + a[1]) / d[2].first;
    }
  }
  printf("%f\n", ans);
  return 0;
}